# Is there a Boltzmann distribution for a system with continuous energy?

• I
• Old Person
In summary, the Boltzman distribution is a limit of the quantum (Bose-Einstein or Fermi-Dirac) distributions, and is used to calculate the probability of finding a system in a given energy state.
Old Person
TL;DR Summary
For a system that can take a discrete set of states with energy E_n and it's assumed to be at (or in equillibrium with a heat bath at) temperature T, the Boltzmann distribution will give the probability of it being in state E_n. What if the system was assumed to have a continuous range of energies possible, i.e. as if there were an infinite set of microstates and no discrete gaps between the energies it can have?
Hi.

I'm not sure where to put this question, thermodynamics or the quantum physics forum (or somewhere else).

For a system in equillibrium with a heat bath at temperature T, the Boltzman distribution can be used.
We have the probability of finding the system in state n is given by ##p_n = \frac{e^{-E_n/KT}}{Z} ##
with K = Boltzmann constant, Z = partition function (as usual).

That's perfectly fine when the system has discrete energy states. We can calculate Z which is a (possibly inifnite but countable) sum of terms etc.

Now suppose that the system does not have a discrete set of energy states but instead the energy of the system could be any positive Real number, it is continuous. Is there a similar formula, presumably using integrals, to determine the probability of finding the system with an amount of Energy between some upper and lower limits?

More details can be provided but I'm not sure I need to take up more of your time reading anything. Thank you for your time.

The Boltzmann distribution is the classical limit of the quantum (Bose-Einstein or Fermi-Dirac) distributions, and indeed the socalled "thermodynamical limit", i.e., for a gas the limit ##V \rightarrow \infty##, while keeping the densities constant, is not that trivial. The outcome is that the probabilities for the discrete case becomes a single-particle phase-space density function, i.e., the number of particles in a phase-space volume element ##\mathrm{d}^3 x \mathrm{d}^3 p## for an ideal gas is given by
$$\frac{\mathrm{d}^6 N}{\mathrm{d}^3 x \mathrm{d}^3 p} = \frac{g}{(2 \pi \hbar)^3} \exp \left (-\frac{E(\vec{p})-\mu}{k_{\text{B}} T} \right),$$
where ##g## is the degneracy factor (counting, e.g., spin-degrees of freedom), ##T## the temperature, and ##\mu## the chemical potential.

Old Person
It will take me a while to study the physics you are describing (maybe next year for example) but I think I have got the general idea.

There is no simple derivation of a formula (analagous to the Boltzmann distribution) that seems to apply. (and I should look at the general theory of quantum distributions when I can).

Please correct me if that's wrong - but otherwise thank you very much for your time and attention.

Every time the system is at one of the possible states. For this reason, the sum of all probabilities should be equal to unity. Or the integral over the whole range of energies, in your example. You should find out, what is the normalization of the distributions.

Old Person
Thanks @napdmitry.
I've more or less answered or thought about what I needed to do now anyway - but thanks for your time. I may not be following this thread as much in the future.

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